Simplicial chain maps carried by specified cones are chain homotopic

Set $h_{-1}=0$. For an oriented vertex $s$, $z=f(s)-g(s)$ has augmentation $1-1=0$. It lies in its carrier, so $h_0(s)=c_{s}z$ satisfies $\mathrm{\partial }{h}_0(s)=z$, using $\mathrm{\partial }{c}_s+c_s\mathrm{\partial }=1$. Changing the sign of the generator changes $h_0$ by the same sign.

Suppose $h$ is defined through degree $n-1$ with $\mathrm{\partial }h+h\mathrm{\partial }=f-g$ there. For an oriented $n$-simplex $s$ put $z=f(s)-g(s)-h(\mathrm{\partial }s)$. The nesting of the carriers puts all summands in $C_n(\mathrm{\Phi }(s))$. Since $f,g$ commute with boundary, $\mathrm{\partial }z=(f-g)(\mathrm{\partial }s)-\mathrm{\partial }h(\mathrm{\partial }s)=h{\mathrm{\partial }}^{2}s=0$. Here ${\mathrm{\partial }}^2=0$ is part of the given chain complexes.

Define $h_n(s)=c_{s}z$. The contraction identity gives $\mathrm{\partial }{h}_n(s)=z-c_s\mathrm{\partial }z=z$, so $\mathrm{\partial }{h}_n(s)+h_{n-1}(\mathrm{\partial }s)=f(s)-g(s)$. The formula is alternating in the original oriented representative: the boundary and $f,g$ are alternating, the already defined $h$ is linear, and $c_s$ depends only on the underlying face. It therefore defines a homomorphism without choosing orientations on all simplices. Induction defines all degrees, carried by $\mathrm{\Phi }$, with the asserted identity; in degree $-1$ both maps are the identity on $\mathbb{Z}$, so their difference is zero. This is a chain homotopy by definition.